Explicit Description of the Zassenhaus Formula

We explicitly describe an expansion of $e^{A+B}$ as an infinite sum of the products of $B$ multiplied by the exponential function of $A$. This is the explicit description of the Zassenhaus formula. We also express the Baker-Campbell-Hausdorff formula in a different manner.


Introduction
In various topics in physics and mathematics, we often have to expand the exponential function of two operators A and B such as e A+B in a certain situation (for instance, [1] and [2]). An expansion is described as the Zassenhaus formula ( [3] and references therein): Unfortunately, however, the above two expressions are rather complicated because we sequentially obtain the explicit expression of higher order terms in the operators A and B. In this paper, we will obtain a new description of the Zassenhaus formula in which all of the higher order terms are explicitly expressed.

Derivation
First of all, we expand (A + B) n and move all the operator A to the right in each term, and define the following expression: where X m are polynomials involving B l , commutators [A, [A, · · · [A, B]]], and their multiplications. By using X m , we obtain the exponential function of A + B as the following form: There exists a recursion relation among X m in such a way that where L A O is the commutator between A and a certain operator O such as It is easy to derive (5) when we compute (A + B) n+1 as the product of (A + B) and (A + B) n in terms of (3). Let us evaluate the relation (5). It is convenient to express X m as the sum of new polynomials X m,p : where p denotes the power of B in X m,p , whose examples can be seen in the appendix. Substituting the expression (6) into the relation (5), we find three recursion relations We immediately obtain the solutions of these relations (the proof is exhibited in the appendix): Here we introduced the terminology B m defined the above. The solution (8c) can be described in an explicit way if we iteratively use (8c) until we reach X m−(k1+...+kp−1),1 = B m−(k1+...+kp−1) given by (8a). Hence we obtain For simplicity, we further introduced the description B m ≡ 1 m! B m . Applying (9) to (3) and (6), we obtain the explicit expansion of e A+B in terms of the products of B m : Relabeling k i and m − (k 1 + . . . + k p−1 ) to n i and n p respectively, we obtain the final form n p · · · n 1 n p (n p + n p−1 ) · · · (n p + . . . + n 1 ) We have a comment that each k i , as well as the new label n i , is unbounded from above because m goes to infinity. It turns out that (11) is the the explicit description of the Zassenhaus formula (2) without using the functions Z n . We understand that the exponential form e B in the right-hand side of (2) can be obtained from (11) when we extract the terms of the products only of B 1 = B. However, it is hard to extract e Zm of arbitrary m from (11). Because of the iterative definition of Z m in (2), we have to obtain the explicit expression of all Z l where l ≤ m − 1 beforehand to determine Z m .
Furthermore, if we transpose (11) and rename A T and B T to A and B, we obtain (−1) (np+...+n1)−p n p · · · n 1 n p (n p + n p−1 ) · · · (n p + . . . This is the explicit description of (1) without using the functions Z n . We should notice that the ordering of the operators B ni is different from that of (11).
The descriptions we obtained are quite useful if the product of the operator B ni is truncated at a certain level such as B n k B n k−1 · · · B n1 = 0, which originates from the nilpotency of the operator B of degree k, i.e., B k = 0.

Baker-Campbell-Hausdorff formula
We can also discuss the Baker-Campbell-Hausdorff (BCH, for short) formula by using the descriptions (11) and (12), though the general form (13) has already been well-known (see, for instance, [4]). We would like to use the operator e Z rather than Z, because we often encounter the exponential form such as e X e Y in quantum mechanics. Multiplying (11) by e −A from the right and replacing A + B and −A with X and Y respectively, we obtain e X e Y = 1 + ∞ p=1 ∞ n1,...,np=1 (−1) (np+...+n1)−p n p · · · n 1 n p (n p + n p−1 ) · · · (n p + . . . + n 1 ) X np · · · X n1 , (14a) On the other hand, multiplying (12) by e −A from the left and replacing −A and A + B with X and Y respectively, we find n p · · · n 1 n p (n p + n p−1 ) · · · (n p + . . . The original BCH formula Z = log(e X e Y ) consists only of X, Y and their commutators. On the other hand, we immediately find that powers of X and Y directly contribute to the expansion in both (14) and (15). These two seem to be different feature. However, the powers of X and Y in (14) and (15) originate from the corresponding power of X + Y in the Taylor expansion of (13).
We recognize that the operators X and Y in (14) and (15) do not appear on equal footing with each other. In order to describe an expression on equal footing, we simply sum up (14) and (15), and divide it by two. For instance, we evaluate this up to cubic powers of the operators X and Y in such a way that This coincides with the Taylor expansion of e X e Y in the form of (13). Indeed, we can find the coincidence in any powers of the operators. Once we establish the above new descriptions, it would be interesting to apply it to a generalization of the BCH formula such as e X e Y e Z developed in [5].
Here we prove (8c) as the solution of the recursion relation (7c) by mathematical induction. Let us assume that each X k,l with 1 ≤ l ≤ k ≤ m satisfies the expression (8c). We compute L A X m,p + BX m,p−1 : The final form is nothing but X m+1,p . Hence we proved that (8c) is the solution of the relation (7c).